126° Вправа 126Алгебра 8 клас Тарасенкова 2025 · § 3 · стор. 39

126. Зведіть до спільного знаменника дроби.

1) ${\displaystyle \frac{1}{x-4}}$ і ${\displaystyle \frac{1}{2x-8}}$;

${\displaystyle 2x-8=}$${\displaystyle 2(x-4)}$;

${\displaystyle \frac{1^{\mathrm{\setminus }2}}{x-4}=}$${\displaystyle \frac{1\cdot 2}{(x-4)\cdot 2}=}$${\displaystyle \frac{2}{2x-8}}$;

${\displaystyle \frac{1}{2x-8}}$ — без змін;

2) ${\displaystyle \frac{1}{5a+1}}$ і ${\displaystyle \frac{1}{4+20a}}$;

${\displaystyle 4+20a=}$${\displaystyle 4(5a+1)}$;

${\displaystyle \frac{1^{\mathrm{\setminus }4}}{5a+1}=}$${\displaystyle \frac{1\cdot 4}{(5a+1)\cdot 4}=}$${\displaystyle \frac{4}{20a+4}}$;

${\displaystyle \frac{1}{4+20a}=}$${\displaystyle \frac{1}{20a+4}}$;

3) ${\displaystyle \frac{2}{1-3y}}$ і ${\displaystyle \frac{5}{12y-4}}$;

${\displaystyle 12y-4=}$${\displaystyle -4(1-3y)}$;

${\displaystyle \frac{2^{\mathrm{\setminus }4}}{1-3y}=}$${\displaystyle \frac{2\cdot 4}{(1-3y)\cdot 4}=}$${\displaystyle \frac{8}{4-12y}}$;

${\displaystyle \frac{5}{12y-4}=}$${\displaystyle \frac{5}{-(4-12y)}=}$${\displaystyle \frac{-5}{4-12y}}$.